Integrating leads to
in which M i is the initial mass of the rocket and M f its final mass. Evaluating the
integrals then gives
(second rocket equation)
(9-88)
for the increase in the speed of the rocket during the change in mass from M i to
M f . (The symbol “ln” in Eq. 9-88 means the natural logarithm.) We see here the
advantage of multistage rockets, in which M f is reduced by discarding successive
stages when their fuel is depleted. An ideal rocket would reach its destination
with only its payload remaining.
v f Ϫ v i ϭ v rel ln
M i
M f
͵
v f
v i
dv ϭ Ϫv rel ͵
M f
M i
dM
M
,
243
REVIEW & SUM MARY
rocket’s mass. However, M decreases and a increases as fuel
is consumed. Because we want the initial value of a here, we
must use the intial value M i of the mass.
Calculation: We find
(Answer)
To be launched from Earth’s surface, a rocket must have
an initial acceleration greater than
. That is, it
must be greater than the gravitational acceleration at the
surface. Put another way, the thrust T of the rocket engine
must exceed the initial gravitational force on the rocket,
which here has the magnitude M i g, which gives us
(850 kg)(9.8 m/s
2
) ϭ 8330 N.
Because the acceleration or thrust requirement is not met
(here T ϭ 6400 N), our rocket could not be launched from
Earth’s surface by itself; it would require another, more
powerful, rocket.
g ϭ 9.8 m/s
2
a ϭ
T
M i
ϭ
6440 N
850 kg
ϭ 7.6 m/s
2
.
Sample Problem 9.09 Rocket engine, thrust, acceleration
In all previous examples in this chapter, the mass of a system
is constant (fixed as a certain number). Here is an example of
a system (a rocket) that is losing mass. A rocket whose initial
mass M i is 850 kg consumes fuel at the rate
The
speed v rel of the exhaust gases relative to the rocket engine is
2800 m/s.What thrust does the rocket engine provide?
KEY IDEA
Thrust T is equal to the product of the fuel consumption
rate R and the relative speed v rel at which exhaust gases are
expelled, as given by Eq. 9-87.
Calculation: Here we find
(Answer)
(b) What is the initial acceleration of the rocket?
KEY IDEA
We can relate the thrust T of a rocket to the magnitude a of
the resulting acceleration with
, where M is the
T ϭ Ma
ϭ 6440 N Ϸ 6400 N.
T ϭ Rv rel ϭ (2.3 kg/s)(2800 m/s)
R ϭ 2.3 kg/s.
Additional examples, video, and practice available at WileyPLUS
Center of Mass The center of mass of a system of n particles is
defined to be the point whose coordinates are given by
(9-5)
or
(9-8)
where M is the total mass of the system.
r
:
com ϭ
1
M ͚
n
iϭ1
m i r
:
i ,
x com ϭ
1
M ͚
n
iϭ1
m i x i , y com ϭ
1
M ͚
n
iϭ1
m i y i , z com ϭ
1
M ͚
n
iϭ1
m i z i ,
Review & Summary
Newton’s Second Law for a System of Particles The
motion of the center of mass of any system of particles is governed
by Newton’s second law for a system of particles, which is
.
( 9 - 1 4 )
Here
is the net force of all the external forces acting on the sysF
:
net
F
:
net ϭ M a
:
com
tem, M is the total mass of the system, and
is the acceleration
of the system’s center of mass.
a
: com
in which M i is the initial mass of the rocket and M f its final mass. Evaluating the
integrals then gives
(second rocket equation)
(9-88)
for the increase in the speed of the rocket during the change in mass from M i to
M f . (The symbol “ln” in Eq. 9-88 means the natural logarithm.) We see here the
advantage of multistage rockets, in which M f is reduced by discarding successive
stages when their fuel is depleted. An ideal rocket would reach its destination
with only its payload remaining.
v f Ϫ v i ϭ v rel ln
M i
M f
͵
v f
v i
dv ϭ Ϫv rel ͵
M f
M i
dM
M
,
243
REVIEW & SUM MARY
rocket’s mass. However, M decreases and a increases as fuel
is consumed. Because we want the initial value of a here, we
must use the intial value M i of the mass.
Calculation: We find
(Answer)
To be launched from Earth’s surface, a rocket must have
an initial acceleration greater than
. That is, it
must be greater than the gravitational acceleration at the
surface. Put another way, the thrust T of the rocket engine
must exceed the initial gravitational force on the rocket,
which here has the magnitude M i g, which gives us
(850 kg)(9.8 m/s
2
) ϭ 8330 N.
Because the acceleration or thrust requirement is not met
(here T ϭ 6400 N), our rocket could not be launched from
Earth’s surface by itself; it would require another, more
powerful, rocket.
g ϭ 9.8 m/s
2
a ϭ
T
M i
ϭ
6440 N
850 kg
ϭ 7.6 m/s
2
.
Sample Problem 9.09 Rocket engine, thrust, acceleration
In all previous examples in this chapter, the mass of a system
is constant (fixed as a certain number). Here is an example of
a system (a rocket) that is losing mass. A rocket whose initial
mass M i is 850 kg consumes fuel at the rate
The
speed v rel of the exhaust gases relative to the rocket engine is
2800 m/s.What thrust does the rocket engine provide?
KEY IDEA
Thrust T is equal to the product of the fuel consumption
rate R and the relative speed v rel at which exhaust gases are
expelled, as given by Eq. 9-87.
Calculation: Here we find
(Answer)
(b) What is the initial acceleration of the rocket?
KEY IDEA
We can relate the thrust T of a rocket to the magnitude a of
the resulting acceleration with
, where M is the
T ϭ Ma
ϭ 6440 N Ϸ 6400 N.
T ϭ Rv rel ϭ (2.3 kg/s)(2800 m/s)
R ϭ 2.3 kg/s.
Additional examples, video, and practice available at WileyPLUS
Center of Mass The center of mass of a system of n particles is
defined to be the point whose coordinates are given by
(9-5)
or
(9-8)
where M is the total mass of the system.
r
:
com ϭ
1
M ͚
n
iϭ1
m i r
:
i ,
x com ϭ
1
M ͚
n
iϭ1
m i x i , y com ϭ
1
M ͚
n
iϭ1
m i y i , z com ϭ
1
M ͚
n
iϭ1
m i z i ,
Review & Summary
Newton’s Second Law for a System of Particles The
motion of the center of mass of any system of particles is governed
by Newton’s second law for a system of particles, which is
.
( 9 - 1 4 )
Here
is the net force of all the external forces acting on the sysF
:
net
F
:
net ϭ M a
:
com
tem, M is the total mass of the system, and
is the acceleration
of the system’s center of mass.
a
: com